1. For which one of the following values of the equilibrium constant does the re

Question

1. For which one of the following values of the equilibrium constant does the reaction mixture at equilibrium contain the greatest fraction of reactants

1e5, 1e3, 1e0, 1e-3, 1e-5

2.

Carbon disulfide and chlorine react according to the following equation:

CS2 (g) + 3 Cl2 (g) S2Cl2 (g) + CCl4 (g)

When 1.00 mol of carbon disulfide and 3.00 mol of chlorine are placed in a 2.00 L container and allowed to come to equilibrium, the mixture is found to contain 0.250 mol of carbon tetrachloride. What is the amount of chlorine at equilibrium? Report your answer in number of moles to 2 s.f. Do not include units in your answer. Do not use scientific notation for your answer.

3. Hydrogen gas and bromine vapor react to form gaseous hydrogen bromide at 1300 K. The value of the equilibrium constant is 1.6 x 105. What is the value of the equilbrium constant for the decomposition of gaseous hydrogen bromide to form hydrogen gas and bromine vapor at 1300 K? Report your answer to 2 s.f. and use scientific notation. If your answer was 123,000 you would report it as 1.2e5.

1e5, 1e3, 1e0, 1e-3, 1e-5

Explanation / Answer

Solution :-

(1) The rate constant of reaction has more value it meant that the rate of reaction is rapid and more products are formed from the reactants and fraction of reactants becomes less, inorder to have greatest fraction of reactants the rate constant must have a low value i.e. 10^-5

       so the answer is E.

(2)

The given equation is

CS2 (g) + 3 Cl2 (g) <------> S2Cl2 (g) + CCl4 (g)

Initial amount of CS2 (g) = 1.0 mol

Initial amount of Cl2 (g) = 3.0 mol

Volume of container , V = 2.0 L

CS2 (g) + 3 Cl2 (g) <------> S2Cl2 (g) + CCl4 (g)

             mol:

              Initial (I):    1.0         3.0                     0                0

          Change (C):        -x        -3x                       +x             +x

   Equilibrium (E): 1.0 -x        3.0-3x                  x               x

But at equilibrium the amount of CCl4 (g) present = 0.250 mol

                                                           so         x = 0.250 mol

Therefore amount of Cl2 (g) present at equilibrium = (3.0 - 3x ) mol

= (3.0 -3*0.250) mol

                                                                                     = 2.25 mol

        So amount of Cl2 (g) present at equilibrium = 2.25 mol

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